How do I sum the series #sum_(n=1)^oon(3/4)^n#?

Question

Emilia Aguilar · Accepted Answer

The answer is: 12.

First of all: "How much is the sum of a geometric series?".

#sum_(n=0)^(+oo)r^n=a_0/(1-r)#, where r (such as #|r|<1#) is called "ratio" (#r=a_k/a_(k-1)#) and #a_0# is the first term of the series (in this case #a_0=r^0=1#).

Let's list all the terms of the series:

#1*3/4 + 2*(3/4)^2+3*(3/4)^3+4*(3/4)^4+...#, but we can also say:

#3/4+9/16+9/16+27/64+27/64+27/64+81/256+81/256+81/256+81/256+...#,

or, better:

#3/4+9/16+27/64+81/256+...+#

#9/16+27/64+81/256+...+#

#27/64+81/256+...+#

#81/256+...+#

#...#

The first sum is:

#sum_(n=1)^(+oo)(3/4)^n#,

The second sum is:

#sum_(n=2)^(+oo)(3/4)^n#,

The third sum is:

#sum_(n=3)^(+oo)(3/4)^n#,

The fourth sum is:

#sum_(n=4)^(+oo)(3/4)^n#,

#...#

So:

#sum_(n=1)^(+oo)n(3/4)^n=sum_(n=1)^(+oo)(3/4)^n+sum_(n=2)^(+oo)(3/4)^n+ sum_(n=3)^(+oo)(3/4)^n+sum_(n=4)^(+oo)(3/4)^n+...=#.

#=3/4*1/(1-3/4)+9/16*1/(1-3/4)+27/64*1/(1-3/4)+81/256*1/(1-3/4)+...=#

#=3/4*4+9/16*4+27/64*4+81/256*4+...=#

#=4(3/4+9/16+27/64+81/256+...)=#

#=4*sum_(n=1)^(+oo)(3/4)^n=4*3/4*1/(1-3/4)=4*3/4*4=12#.

Isabella Ackerman · Answer

undefined To sum the series $ \sum_{n=1}^{\infty} n \left(\frac{3}{4}\right)^n $, you can use the formula for the sum of an infinite geometric series:

$$ S = \frac{a}{1 - r} $$

where $ a $ is the first term of the series and $ r $ is the common ratio.

In this series, the first term ($ a $) is $ \frac{3}{4} $, and the common ratio ($ r $) is also $ \frac{3}{4} $ because each term is obtained by multiplying the previous term by $ \frac{3}{4} $.

Thus, plugging these values into the formula:

$$ S = \frac{\frac{3}{4}}{1 - \frac{3}{4}} $$

$$ S = \frac{\frac{3}{4}}{\frac{1}{4}} $$

$$ S = 3 $$

So, the sum of the series $ \sum_{n=1}^{\infty} n \left(\frac{3}{4}\right)^n $ is 3.